Suppose that f(x) = \frac{\ln(x)}{x^{20}}. Find f'(2).

Question:

Suppose that {eq}f(x) = \frac{\ln(x)}{x^{20}} {/eq}. Find {eq}f'(2). {/eq}

Quotient Rule:

Let {eq}t(x) {/eq} and {eq}g(x) {/eq} be functions of {eq}x {/eq} then the derivative of {eq}\frac{{t(x)}}{{g(x)}} {/eq} is given by:

{eq}{\,\frac{d}{{dx}}\frac{t}{g} = \frac{{g\frac{{dt}}{{dx}} - t\frac{{dg}}{{dx}}}}{{{g^2}}}} {/eq}

The following rules are relevant to this problem:

1.{eq}\frac{r}{r} = 1 {/eq}

2.{eq}{\frac{d}{{dx}}\ln x = \frac{1}{x},\frac{d}{{dx}}{x^n} = n{x^{n - 1}}} {/eq}

Given that: {eq}\displaystyle f(x) = \frac{{\ln (x)}}{{{x^{20}}}} {/eq}

{eq}\displaystyle \eqalign{ & f(x) = \frac{{\ln (x)}}{{{x^{20}}}} \cr & {\text{Differentiating }}f(x){\text{ w}}{\text{.r}}{\text{.t}}{\text{. '}}x{\text{';}} \cr & \frac{{df}}{{dx}} = \frac{d}{{dx}}\frac{{\ln (x)}}{{{x^{20}}}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{{x^{20}}\frac{d}{{dx}}\ln x - \ln x\frac{d}{{dx}}{x^{20}}}}{{{x^{40}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{From quotient rule}}} \right) \cr & \,\,\,\,\,\,\,\,\, = \frac{{{x^{20}}\left( {\frac{1}{x}} \right) - \ln x\left( {20{x^{20 - 1}}} \right)}}{{{x^{40}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Chain rule,}}\frac{d}{{dx}}\ln x = \frac{1}{x},\frac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right) \cr & \,\,\,\,\,\,\,\,\, = \frac{{{x^{19}} - 20\ln x\left( {{x^{19}}} \right)}}{{{x^{40}}}} \cr & {\text{Cancel the common term;}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{1 - 20\ln x}}{x} \cr & \cr & f'\left( x \right) = \frac{{1 - 20\ln x}}{x} \cr & f'\left( x \right){\text{ at }}x = 2; \cr & f'\left( 2 \right) = \frac{{1 - 20\ln 2}}{2} \cr} {/eq}